Question

CMR với mọi A, B, C > 0

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)

Answer 1

Ta có \(\frac{1a^2}{b}+b\ge2a\)

\(\frac{1b^2}{c}+c\ge2b\)

\(\frac{1c^2}{a}+a\ge2c\)

Cộng vế theo vế ta được

\(\frac{1a^2}{b}+\frac{b^2}{C}+\frac{c^2}{a}\)+ a + b + c \(\ge\)2(a + b + c)

<=> \(\frac{1a^2}{b}+\frac{b^2}{C}+\frac{c^2}{a}\)\(\ge\)a + b + c

Answer 2

Dùng Swartz 2 số 1 ra luôn mà